trigonometric

**n7167113** · March 16th 2009, 08:57 PM

I have 2 questions:

find for z ; where z^3=sqrt[imaginary]
Simplify: sin x/(1-cos x)

can somebody help me with these 2 questions showing working out?
Please & Thank You

**Soroban** · March 16th 2009, 09:28 PM

Hello, n7167113!

Find $z\!:\;\;z^3\:=\:\sqrt{i}$

We have: . $z \:=\:i^{\frac{1}{6}}$

We know: . $i \:=\:\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}$ ,

Use DeMoivre's Formula for the six roots of: . $\left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right)^{\frac{1}{6}}$

Simplify: . $\frac{\sin x}{1-\cos x}$

This doesn't simplify very much, but here goes . . .

Multiply by $\frac{1 +\cos x}{1+\cos x}\!:\;\;\frac{\sin x}{1 - \cos x}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{\sin x(1 + \cos x)}{1-\cos^2\!x} \;=\;\frac{\sin x(1 + \cos x)}{\sin^2\!x}$

. . $= \;\frac{1+\cos x}{\sin x} \;=\;\frac{1}{\sin x} + \frac{\cos x}{\sin x} \;=\;\csc x + \cot x$

**coolguy99** · March 16th 2009, 09:33 PM

For the simplification one, try multiplying both the numerator and denominator by (1+cos(x)), and from there you should be able to get a nice answer

edit> I was beat to the punch

Thread: trigonometric

LinkBack

Thread Tools

Display

trigonometric

Similar Math Help Forum Discussions

Trigonometric sum.

Applications of Derivatives of Trigonometric and Inverse Trigonometric Functions

Trigonometric Functions and Inverse Trigonometric Functions

Trigonometric Sum

Trigonometric Sum

Search Tags